Question

Suppose you have a pendulum clock which keeps correct time on Earth(acceleration due to gravity = 1.6 m/s2). For ever hour interval (on Earth) the Moon clock:

A) (9.8/1.6)h
B) 1 h
C) the square root of 9.8/1.6 h
D) (1.6/9.8)h
E) the square root of 1.6/9.8 h

Answer 1

The time-period of a simple pendulum is

Time =  2 π √(length/grav-accel)

After unraveling the question, then completing it, and working out what I believe it's trying to ask, the choice that correctly answers the question that I have invented is choice-E .

Answer 2

The simple pendulum allows to find the correct answer for the relationship between the time intervals of an earth clock that is carried to the moon is:

         E) The factor is: $\frac{T_{earth} }{ T_{moon} } = \sqrt{\frac{1.6}{9.8} }$  

The simple pendulum is a system formed by a rope and a weight is its end, the angular velocity of this system is given by the expression.

           $w = \sqrt{\frac{g}{l} }$  

Where w is the angular velocity, g the acceleration of gravity and l the length of the pendulum.

They indicate that the clock tells the correct time on the earth that has a gravity of g earth = 9.8 m / s² and ask for each time interval how much this clock marks if it were on the Moon that has gravity of g = 1.6 m / s².

Angular velocity and period are related.

          $w = \frac{2\pi }{T}$  

Let's substitute.

          $T = 2 \pi \ \sqrt{ \frac{l}{g} }$  

Let's write this expression for the two places.

Earth.

          $T_{earth} = 2 \pi \sqrt{ \frac{l}{g_{earth}} }$  

Moon

          $T_{moon} = 2 \ pi \ \sqrt{ \frac{l}{g_{moon}}$  

Let's look for the relationship between these two quantities.

         $\frac{T_{earth}}{T_{moon}} = \sqrt{ \frac{g_{moon}}{g_{earth}} } \\\frac{T_{earth}}{T_{moon}} = \sqrt{ \frac{1.6}{9.8} } \\\frac{T_{earth}}{T_{moon}} = 0.404$

Let's analyze the different answers:

A) False. The dependency is with the square root.

B) False the change factor is 0.404.

C) False. The acceleration of gravity is reversed.

D) False. The relationship is with the square root.

E) True. You agree with the answer you will find.

In conclusion, using the simple pendulum we can find the correct answer for the relationship between the time intervals of an earth clock that is carried to the moon is:

         E) The factor is: $\frac{T_{earth}}{ T_[moon}} = \sqrt{ \frac{1.6}{9.8} }$  

Learn more about the simple pendulum here: brainly.com/question/14759840